Chemistry: An Atoms First Approach
Chemistry: An Atoms First Approach
2nd Edition
ISBN: 9781305079243
Author: Steven S. Zumdahl, Susan A. Zumdahl
Publisher: Cengage Learning
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Chapter 12, Problem 78AE

(a)

Interpretation Introduction

Interpretation: The pressure of PCl5 in undissociated form is to be calculated. The partial pressure, the total pressure and the percent dissociation of PCl5 at equilibrium is to be calculated.

Concept introduction: The equilibrium constant Kp describes the ratio of the reactant to the product on the equilibrium conditions in terms of the partial pressure..

At equilibrium sum of all the partial pressure is equal to the total pressure.

Law of mass action is applicable on the equilibrium reactions.

The molar mass of PCl5=208.23gmol-1 .

Dissociation is also known as degree of dissociation which tells that dissociated amount of the species at a particular time.

To determine: The pressure of PCl5 in undissociated form.

(a)

Expert Solution
Check Mark

Answer to Problem 78AE

Answer

The pressure of PCl5 in undissociated form is 1.16atm_ .

Explanation of Solution

Explanation

The pressure of PCl5 in undissociated form is 1.16atm_ .

Given

The reactions are given as,

PCl5(s)PCl5(g)+Cl2(g)

The equilibrium constant Kp=11.5 at 600K .

Mass of PCl5=2.450g .

Volume of the PCl5 =500ml .

Moles of the PCl5 is determined by the formula,

n=mM

Where,

  • n is the number of moles.
  • m is the given mass.
  • M is the molar mass.

Substitute the values in the above equation.

n=mM=2.450g208.23gmol-1=0.01176mol

Volume of the PCl5 gas in Liter =0.50L

The pressure of PCl5 in undissociated form is given by the formula,

P=(nV)RT

Where,

  • P is the pressure.
  • V is the volume in L.
  • R is the gas constant.
  • T is the temperature.

Substitute the values in the above equation.

P=(nV)RT=(0.01176mol0.50L)0.08206LatmK1mol-1×600K=1.16atm_

Conclusion

The pressure of PCl5 in undissociated form is 1.16atm_ .

(b)

Interpretation Introduction

Interpretation: The pressure of PCl5 in undissociated form is to be calculated. The partial pressure, the total pressure and the percent dissociation of PCl5 at equilibrium is to be calculated.

Concept introduction: The equilibrium constant Kp describes the ratio of the reactant to the product on the equilibrium conditions in terms of the partial pressure..

At equilibrium sum of all the partial pressure is equal to the total pressure.

Law of mass action is applicable on the equilibrium reactions.

The molar mass of PCl5=208.23gmol-1 .

Dissociation is also known as degree of dissociation which tells that dissociated amount of the species at a particular time.

To determine: The partial pressure of PCl5 at equilibrium.

(b)

Expert Solution
Check Mark

Answer to Problem 78AE

Answer

The partial pressure of PCl5 at equilibrium is 0.10atm_ .

Explanation of Solution

Explanation

The partial pressure of PCl5 at equilibrium is 0.10atm_ .

The initial partial pressure of PCl5 is calculated in step 1 =1.16atm and the change in partial pressure is x .

The (ICE-chart) that is the initial pressure, change in pressure and equilibrium pressure is given as,

PCl5(g)PCl3(g)Cl2(g)Initialconcentration1.1600Changeinconcentration-x+x+xEquilibriumconcentration1.16-xxx

The equilibrium constant in terms of partial pressure for the given reaction is given by the formula.

kp=PPCl3PCl2PPCl5

Where,

  • PPCl3 is the partial pressure of PCl3 .
  • PCl2 is the partial pressure of Cl2 .
  • PPCl5 is the partial pressure of PCl5 .

Substitute the values of partial pressures from the ICE-table in the above equation.

kp=PPCl3PCl2PPCl511.5=(x)(x)(1.16x)x=1.06

From the ICE table the equilibrium partial pressure of PCl5(g) is calculated as,

PPCl5=1.16x=1.161.06=0.10atm_

Conclusion

The partial pressure of PCl5 at equilibrium is 0.10atm_ .

(c)

Interpretation Introduction

Interpretation: The pressure of PCl5 in undissociated form is to be calculated. The partial pressure, the total pressure and the percent dissociation of PCl5 at equilibrium is to be calculated.

Concept introduction: The equilibrium constant Kp describes the ratio of the reactant to the product on the equilibrium conditions in terms of the partial pressure..

At equilibrium sum of all the partial pressure is equal to the total pressure.

Law of mass action is applicable on the equilibrium reactions.

The molar mass of PCl5=208.23gmol-1 .

Dissociation is also known as degree of dissociation which tells that dissociated amount of the species at a particular time.

To determine: The total pressure in the bulb at equilibrium.

(c)

Expert Solution
Check Mark

Answer to Problem 78AE

Answer

The total pressure of PCl5 at equilibrium is 2.22atm_ .

Explanation of Solution

Explanation

The total pressure of PCl5 at equilibrium is 2.22atm_ .

At equilibrium the partial pressure is equal to the sum of the individual pressure of the all gases. Therefore,

Ptotal=Ppartial

Where,

  • Ptotal is the total pressure at equilibrium.
  • Ppartial is the partial pressure of the gases.
  • represents the sum of all partial pressures.

Substitute the given value of total pressure and value of partial pressure of all gases.

Ptotal=Ppartial=PPCl5+PPCl3+PCl2=0.10+1.062+1.062=2.22atm_

Conclusion

The total pressure of PCl5 at equilibrium is 2.22atm_ .

(d)

Interpretation Introduction

Interpretation: The pressure of PCl5 in undissociated form is to be calculated. The partial pressure, the total pressure and the percent dissociation of PCl5 at equilibrium is to be calculated.

Concept introduction: The equilibrium constant Kp describes the ratio of the reactant to the product on the equilibrium conditions in terms of the partial pressure..

At equilibrium sum of all the partial pressure is equal to the total pressure.

Law of mass action is applicable on the equilibrium reactions.

The molar mass of PCl5=208.23gmol-1 .

Dissociation is also known as degree of dissociation which tells that dissociated amount of the species at a particular time.

To determine: The percent dissociation of PCl5 at equilibrium.

(d)

Expert Solution
Check Mark

Answer to Problem 78AE

Answer

The percent dissociation of PCl5 at equilibrium is 91.4%_ .

Explanation of Solution

Explanation

The percent dissociation of PCl5 at equilibrium is 91.4%_ .

The initial moles of PCl5 calculated in step 1 =0.0117mol

At equilibrium the number of moles calculated by the formula,

n=PVRT

Substitute the values in the above equation.

n=PVRT=(0.10atm)(0.50L)0.08206LatmK-1mol-1×600K=0.001mol

The number of reacted moles is given as,

=0.01170.001=0.0107mol

Therefore, the percent degree of dissociation of PCl5 at equilibrium is given by the formula,

%α=nreactedninitial×100

Where,

  • %α is the percent degree of dissociation.
  • nreacted is the number of reacted moles.
  • ninitial is the initial number of moles.

Substitute the values in the above equation.

%α=nreactedninitial×100=0.01070.0117×100=91.4%_

Conclusion

The percent dissociation of PCl5 at equilibrium is 91.4%_ .

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Chapter 12 Solutions

Chemistry: An Atoms First Approach

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